Giải phương trình :  $2 \cos ^2 \frac{ 6x}{ 5} +1 = 3 \cos \frac{ 8x}{ 5}      (1)$
Đặt  $t= \frac{ 4x}{ 5} $
Ta được $\cos \frac{ 8x}{ 5} = \cos 2t  ,  2 \cos ^2 x \frac{6x }{5 }= 1 + \cos \frac{12x }{ 5} = 1+\cos 3t$
Vậy phương trình $(1)$ có dạng:
                $1+\cos 3t +1= 3 \cos 2t$
          $\Leftrightarrow 2+4 \cos ^3 t - 3 \cos t = 3(2 \cos ^2 t -1)$
         
          $\Leftrightarrow 4 \cos ^3 t - 6 \cos ^2 t - 3 \cos t +5=0$
          $\Leftrightarrow \left[ \begin{array}{l}
\cos t  = 1\\
\cos t   = \frac{{1 + \sqrt {21} }}{4} > 1  ( loại )\\
\cos t  = \frac{{1 - \sqrt {21} }}{4}
\end{array} \right.$
Vậy phương trình $(1)$ có nghiệm của phương trình :
          $\left[ \begin{array}{l}
\cos \frac{{4x}}{5} = 1\\
\cos \frac{{4x}}{5} = \frac{{1 - \sqrt {21}   }}{4}=cos \alpha
\end{array} \right.    \Leftrightarrow    \left[ \begin{array}{l}
\frac{{4x}}{5} = 2k\pi \\
\frac{{4x}}{5} =
\pm \alpha  + 2k\pi
\end{array} \right.   \Leftrightarrow    \left[ \begin{array}{l}
x = \frac{{5k\pi }}{2}\\
x =  \pm \frac{{5\alpha }}{4} + \frac{{5k\pi }}{2}
\end{array} \right.$

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